**Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.6**

**Determinants Exercise 4.6 Solutions**

**1. Examine the consistency of the system of equations. x + 2y = 22x + 3y = 3**

**Solution**

The given system of equation is :

x + 2y = 2

2x + 3y = 3

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 1(3) - 2(2) = 3 - 4 = -1 ≠ 0

∴ A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**2. Examine the consistency of the system of equations.2 x − y = 5x + y = 4**

**Solution**

The given system of equations is :

2x - y = 5

x + y = 4

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 2|1| - (-1)(1) = 2 + 1 = 3 ≠ 0

∴ A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**3. Examine the consistency of the system of equations. x + 3y = 52x + 6y = 8**

**Solution**

The given system of equations is :

x + 3y = 5

2x + 6y = 8

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 1(6) - 3(2) = 6 - 6 = 0

∴ A is a singular matrix.

Now,

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

**4. Examine the consistency of the system of equations. x + y + z = 12x + 3y + 2z = 2ax + ay + 2az = 4**

**Solution**

The given system of equations is :

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

This system of equations can be written in the form AX = B, where

Now,

|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)

= 4a - 2a - a = 4a - 3a = a ≠ 0

∴ A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**5. Examine the consistency of the system of equations.3 x − y − 2z = 22y − z = −13x − 5y = 3**

**Solution**

The given system of equations is :

3x - y - 2z = 2

2y - z = - 1

3x - 5y = 3

This system of equations can be written in the form of AX = B, where

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

**6. Examine the consistency of the system of equations.5 x − y + 4z = 52x + 3y + 5z = 25x − 2y + 6z = −1**

**Solution**

The given system of equations is :

5*x* −* y *+ 4*z* = 5

2*x* + 3*y* + 5*z* = 2

5*x* − 2*y* + 6*z* = −1

This system of equations can be written in the form of AX = B, where

Now,

|A| = 5(18 + 10) + 1(12 - 25) + 4(- 4 - 15)

= 5(28) + 1(-13) + 4(-19)

= 140 - 13 - 76

= 51 ≠ 0

A is non - singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

**7. Solve system of linear equations, using matrix method.5x + 2y = 47x + 3y = 5**

**Solution**

The given system of equations can be written in the form of AX = B, where

Now, |A| = 15 - 14 = 1 ≠ 0.

Thus, A is non - singular. Therefore, its inverse exists.

Now,

A^{-1} = (1/|A|) [adj A]

**8. Solve system of linear equations, using matrix method.2x – y = –23x + 4y = 3**

**Solution**

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 8 + 3 = 11 ≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

Now,

**9. Solve system of linear equations, using matrix method.4x – 3y = 33x – 5y = 7**

**Solution**

The given system of equation can be written in the form of AX = B, where

Now,

|A| = -20 + 9 = -11≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

Now,

**10.Solve system of linear equations, using matrix method.5 x + 2y = 33x + 2y = 5**

**Solution**

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 10 - 6 = 4 ≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

**11. Solve system of linear equations, using matrix method.2x + y + z = 1x – 2y – z = 3/23y – 5z = 9**

**Solution**

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 2(10 + 3) - 1(-5 - 3) + 0 = 2(13) - 1(-8) = 26 + 8 = 34 ≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

Now,

A_{11} = 13, A_{12} = 5, A_{13} = 3

A_{21} = 8, A_{22} = -10 , A_{23} = -6

A_{31} = 1, A_{32} = 3, A_{33} = -5

**12. Solve system of linear equations, using matrix method. x − y + z = 42x + y − 3z = 0x + y + z = 2**

**Solution**

The given system of equations can be written in the form of AX= B, where

Now,

|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1= 10 ≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

Now,

A_{11} = 4, A_{12} = -5, A_{13} = 1

A_{21} = 2, A_{22} = 0 , A_{23} = -2

A_{31} = 2, A_{32} = 5, A_{33} = 3

**13. Solve system of linear equations, using matrix method.2 x + 3y + 3z = 5x − 2y + z = −43x − y − 2z = 3**

**Solution**

The given system of equations can be written in the form AX = B, where

Now,

|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6)

= 2(5) - 3(-5) + 3(5)

= 10 + 15 + 15

= 40 ≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

Now,

A_{11} = 5, A_{12} = 5, A_{13} = 5

A_{21} = 3, A_{22} = -13 , A_{23} = 11

A_{31} = 9, A_{32} = 1, A_{33} = -7

**14. Solve system of linear equations, using matrix method. x − y + 2z = 73x + 4y − 5z = −52x − y + 3z = 12**

**Solution**

The given system of equations can be written in the form of AX = B, where

Now,

|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)

= 7 + 19 - 22

= 4 ≠ 0

Thus, A is non - singular. Therefore, its inverse exists.

Now,

A_{11} = 7, A_{12} = -19, A_{13} = -11

A_{21} = 1, A_{22} = -1 , A_{23} = -1

A_{31} = -3, A_{32} = 11, A_{33} = 7

**15. If A = **** find A^{−1}. Using A^{−1} solve the system of equations**

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

**Solution**

A =

∴ |A| = 2(- 4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 ≠ 0

Now,

A_{11} = 0, A_{12} = 2, A_{13} = 1

A_{21} = -1, A_{22} = -9 , A_{23} = -5

A_{31} = 2, A_{32} = 23, A_{33} = 13

**16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.**

**Solution**

Let the cost of onions, wheat, and rice per kg be Rs x, Rs y, and Rs z respectively.

Then, the given situation can be represented by a system of equations as :

4x +3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

This system of equations can be written in the form of AX = B, where

|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24)

= 0 + 90 - 40 = 50 ≠ 0

Now,

A_{11} = 0, A_{12} = 30, A_{13} = -20

A_{21} = -5, A_{22} = 0 , A_{23} = 10

A_{31} = 10, A_{32} = -20, A_{33} = 10

x = 5, y = 8 and z = 8

Hence, the cost of onion per kg is Rs. 5, the cost of wheat per kg is Rs. 8 and the cost of rice per kg is Rs. 8.